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Electromagnetic Fields and Waves: Including Electric Circuits

Paul Lorrain, Dale R. Corson

Chapter 9

Electric Fields Vii - all with Video Answers

Educators

Chapter Questions

01:10

Problem 1

A sample of diamond has a density of $3.5 \times 10^{3}$ kilograms/meter $^{3}$ and a polarization of $10^{7}$ coulomb/meter $^{2}$.
(a) Calculate the average dipole moment per atom.
(b) What is the average separation between the centers of positive and negative charge? The carbon nucleus has a charge $+6 e$ and is surrounded by six electrons. The diameter of an atom is of the order of $10^{-10}$ meter.

Aadit Sharma
Aadit Sharma
Numerade Educator
01:22

Problem 2

Show that, in a nonhomogeneous dielectric, if $\rho_{f}=0$, then $\rho_{b}=$ $-\left(\epsilon_{11} / \epsilon_{r}\right) \boldsymbol{E} \cdot \boldsymbol{\nabla} \epsilon_{r}$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
02:42

Problem 3

The polarization $\boldsymbol{P}$ of a given dielectric sphere is uniform. Here is an elegant way of calculating its field, both inside and outside.
Take the sum of the fields of two spherical charges of densities $\rho$ and $-\rho$, displaced by a distance $\Delta z$, one with respect to the other, $\Delta z$ being the distance between the centers of charge of the individual dipoles. Then $\boldsymbol{P}=\rho \Delta z \hat{z}$ if the vector $\boldsymbol{P}$ points in the positive direction of the $z$-axis.
(a) Show that, inside, $\boldsymbol{E}=-\boldsymbol{P} / 3 \boldsymbol{\epsilon}_{0}$ and is therefore uniform.
(b) Show that, outside, $V$ and $\boldsymbol{E}$ are those of a dipole of moment $\boldsymbol{P} \mathscr{V}$ situated at the center of the sphere of volume $\mathcal{V}$.

Keshav Singh
Keshav Singh
Numerade Educator
08:39

Problem 4

Show that, in a nonhomogeneous medium, $\boldsymbol{\nabla}^{2} V+(\boldsymbol{\nabla} V \cdot \boldsymbol{\nabla} \epsilon) / \epsilon=-\rho_{f} / \epsilon$.

Alex Roush
Alex Roush
Numerade Educator
01:45

Problem 5

The surface of a certain sample of dielectric carries a free charge of density of $\sigma_{f}$.
Show that, close to the surface, $E_{\text {air }}=\sigma_{f} /\left[\left(\epsilon_{r}+1\right) \epsilon_{0}\right]$ and $E_{\text {diel }}=$ $-\sigma_{f} /\left(\epsilon_{r}+1\right) \epsilon_{0}$. We have chosen the outward direction as positive. Note that $D_{\text {air }}-D_{\text {diec }}=\sigma_{f}$, as expected. Also, in the dielectric, $\boldsymbol{D}=\epsilon_{\mathrm{0}} \boldsymbol{E}+\boldsymbol{P}$.

Ankur S
Ankur S
Numerade Educator
20:07

Problem 6

If high-energy electrons bombard a block of insulating material such as Lucite, the electrons penetrate the material and remain trapped inside. If one then gives the block a sharp knock with a conducting object, say with a center punch, the electrons escape, leaving a beautiful tree-like design where the plastic has broken down.
In one particular instance, a 0.1-microampere beam bombarded an area of 25 centimeter $^{2}$ of Lucite $(\epsilon,=3.2)$ for 1 second, and essentially all the electrons were trapped about 6 millimeters below the surface in a region about 2 millimeters thick. The block was 12 millimeters thick.
In the following calculations, neglect edge effects and assume a uniform density for the trapped electrons. Assume also that both faces of the Lucite are in contact with grounded conducting plates.
(a) What is the bound charge density in the charged region?
(b) What is the bound charge density at the surface of the Lucite?
(c) Sketch graphs of $D, E, V$ as functions of position inside the dielectric.
(d) Show that the potential at the center of the sheet of charge is about 4 kilovolts.
(e) What is the energy stored in the block? Could the block explode?

Linda Winkler
Linda Winkler
Numerade Educator
02:13

Problem 7

A conducting body $A$, of arbitrary shape, is grounded. Another conducting body $B$, of arbitrary shape and position, is maintained at a potential $V$. In air, the capacitance is $C_{0}$. Show that the capacitance is $\epsilon_{r} C_{0}$. when the bodies are submerged in a large body of dielectric $\epsilon_{r}$.

Shoukat Ali
Shoukat Ali
Other Schools
08:05

Problem 8

Figure $33-4$ shows a section of coaxial line. A dielectric $\epsilon_{r}$ fills the space between the two conductors. From Probs. $6-5$ and $9-7$, the capacitance per unit length $C^{\prime}$ is $2 \pi \epsilon, \epsilon_{0} / \ln \left(R_{2} / R_{1}\right)$. The outer conductor is grounded, and the inner conductor is at the potential $V$.
(a) Calculate the charge per unit length $\lambda$ on the inner conductor.
(b) Show that the bound charge per unit length on the inner surface of the dielectric is $-\lambda\left(1-1 / \epsilon_{r}\right)$.
Thus the net charge per unit length at the radius $R_{1}$ is $\lambda / \epsilon_{r}$. The bound charge per unit length on the outer surface of the dielectric is similarly $+\lambda\left(1-1 / \epsilon_{r}\right)$, and the net charge per unit length at $R_{2}$ is $-\lambda / \epsilon_{r}$.
(c) Show that the volume density of bound charge is zero.
(d) Draw graphs of $D, E$, and $V$ as functions of the radius $r$ from $r=R_{1}$ to $r=R_{2}$, for $V=100$ volts, $R_{1}=1.00$ millimeter, $R_{2}=10.0$ millimeters, $\epsilon_{r}=3.00 .$

Sarah Mccrumb
Sarah Mccrumb
Numerade Educator
06:35

Problem 9

The dielectric of a parallel-plate capacitor has a permittivity that varies as $\epsilon_{r 0}+a x$, where $x$ is the distance from one plate. The area of a plate is $\mathscr{A}$, and their spacing is $s$.
(a) Find the capacitance.
(b) Show that, if $\epsilon$, varies from $\epsilon_{r 0}$ to $2 \epsilon_{r 0}$, then $C$ is $1.44$ times larger than if $a$ were zero.
(c) Find $P$ from the values of $D$ and $E$ for that case.
(d) Deduce the value of $\rho_{b}$.
(e) Now calculate $\rho_{b}$ from the relation given in Prob. 9-2.
(f) Draw curves of $E, \rho_{b}$, and $P$ as functions of $x$ for $\epsilon_{r 0}=3.00$, $a=\epsilon_{r 0} / s, s=1.00$ millimeter when the applied voltage is $1.00$ volt.

Sheh Lit Chang
Sheh Lit Chang
University of Washington
06:35

Problem 10

When the space between the plates of a parallel-plate capacitor is filled with a dielectric $\epsilon$, it has a capacitance of $C$ farads. If the dielectric is replaced by a material whose resistivity $\rho=1 / \sigma$ is much smaller than that of the electrodes, the resistance between the electrodes is $R$ ohms.
(a) Show that $R C=\rho \in$, neglecting edge effects.
(b) Show that this result also applies to cylindrical and spherical capacitors.
(c) Show that this result applies to any pair of electrodes submerged in a medium whose resistivity $\rho$ is much larger than that of the electrodes.

You should be able to show that the field is unaffected by the conductivity, with the above restriction.

One important application of this fact is the electrolytic plotting tank, which is used for plotting electric fields in two, and in some cases three, dimensions.
If the medium occupies only the region between the electrodes, then $R C$ is again equal to $\rho \epsilon$, except that $C$ does not include the fringing field and is therefore smaller than the true capacitance.
(d) The capacitance per unit length between two parallel wires of diameter $d$ and separated by a distance $D$ is $C^{\prime}=\pi \epsilon / \cosh ^{-1}(D / d)$. Find the conductance between parallel wires 10 millimeters in diameter separated by 100 millimeters and submerged in sea water $(\sigma=5)$.

Sheh Lit Chang
Sheh Lit Chang
University of Washington
01:44

Problem 11

See Prob. 9-10. The electrodes of part (c) are initially charged, and discharge through the medium. Show that $J_{f}+\partial D / \partial t=0$.

Keshav Singh
Keshav Singh
Numerade Educator